6 Determinants

6.1 10 Properties of Determinants

For square matrices,

1. $|I|=1$ for any $n\times n$ identiy matrix $I$.

2. Determinant changes sign when two rows or two columns are exchanged. $$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = - \begin{vmatrix} c & d \\ a & b \end{vmatrix} = - \begin{vmatrix} b & a \\ d & c \end{vmatrix}$$ For permutation matrix $P$ involving $r$ row exchanges, $$|P| = 1 \text{ for even } r,\quad |P| = -1 \text{ for odd } r$$

3. The determinant is a linear function of each row separately.
   Scaling a single row of a matrix by $t$, $$\begin{vmatrix} ta & tb \\ c & d \end{vmatrix} = t \begin{vmatrix} a & b \\ c & d \end{vmatrix}$$ Adding a row of one matrix to the corresponding row of another matrix, $$\begin{vmatrix} a+a^{\prime} & b+b^{\prime} \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a^{\prime} & b^{\prime} \\ c & d \end{vmatrix}$$ Note that the addition can only be distributed across a single row. The remaining rows in all the matrices involved must be unchanged. As a general rule, $$|A + B| \neq |A| + |B|$$

4. If any 2 rows in matrix $A$ are equal, then $|A|=0$. $$\begin{vmatrix} a & b \\ a & b \\ c & d\end{vmatrix} = 0$$ Proof. Swapping the 2 identical rows in $A$, the determinant is $-|A|$ (Property 2). However, the matrix $A$ did not change, so $|A| = -|A| \implies |A|=0$.
   Note that this applies if two rows are scalar multiples of each other. $$\begin{vmatrix} a & b \\ ta & tb \end{vmatrix} = t \begin{vmatrix} a & b \\ a & b \end{vmatrix} = 0$$

5. Adding/Subtracting a multiple of one row to another row does not change $|A|$. $$\begin{vmatrix} a & b \\ c-la & d-lb \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix}$$ for $l$ a scalar.
   Proof. $$\begin{align*} \begin{vmatrix} a & b \\ c-la & d-lb \end{vmatrix} &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a & b \\ -la & -lb \end{vmatrix} \\ &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} - l \begin{vmatrix} a & b \\ a & b \end{vmatrix} \\ &= \begin{vmatrix} a & b \\ c & d \end{vmatrix} - l(0) \end{align*}$$ Note that when performing Gaussian elimination on $A$ to get $U$ (upper triangular; REF form), we either subtract/add multiples of one row to another (does not affect $|A|$), or swap rows (changes sign of $|A|$). Hence $|A| = \pm |U|$.
   Note further that performing further elimination to RREF does change $|A|$, as we would have to scale rows (to make the pivots = 1).

6. If matrix $A$ has a row of zeroes, $|A|=0$. $$\begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix} = 0$$ Proof. $$|A| = \begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix} = \begin{vmatrix} 0 + c & 0 + d \\ c & d \end{vmatrix} = \begin{vmatrix} c & d \\ c & d \end{vmatrix} = 0$$

7. If matrix $A$ is triangular, its determinant is simply the product of its main diagonal entries. $$|A| = a_{11}a_{22} \dots a_{nn}$$ For purely diagonal matrix, each diagonal element can be factored out row by row until you are left with the identity matrix. $$\begin{vmatrix} a_{11} & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{nn} \end{vmatrix} = a_{11} \begin{vmatrix} 1 & 0 & 0 \\ 0 & a_{22} & 0 \\ 0 & 0 & a_{nn} \end{vmatrix} = a_{11}a_{22} \dots a_{nn} \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$ Since $|I| = 1$, then we get the result above.
   For upper triangular matrix $U$, $$|U| = \begin{vmatrix} a & b \\ 0 & d \end{vmatrix}$$ The off-diagonal entries can be made 0 by row elimination. For example, applying $R_1 \leftarrow R_1 - \frac{b}{d}R_2$ to $U$, $$|U| = \begin{vmatrix} a & 0 \\ 0 & d \end{vmatrix} = ad$$ If any diagonal element $a_{ii} = 0$, the elimination process will produce an entire row of zeros. Following Property 6, this means the determinant is $0$, and the matrix is classified as singular.

8. $A$ is singular $\iff |A| = 0$, and $A$ is invertible $\iff |A| \neq 0$.
   If $A$ is singular, the elimination process produces $U$ with a row of zeroes $\implies |U| = 0 \implies |A| = 0$.
   If $A$ is invertible, $U$ will have non-zero pivots along its entire main diagonal. From Rule 7, the product of these pivots is non-zero, hence $|A| = \pm |U|$.
   For a $2 \times 2$ matrix, and $a \neq 0$, $$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ 0 & d - (c/a)b \end{vmatrix} = a \times \left(d - \frac{cb}{a}\right) = ad - bc$$

9. Determinant of product $|AB| = |A||B|$.
   Proof. Consider the function $D(A) = \frac{|AB|}{|B|}$, if it passes these 3 rules, then $D(A)$ is a determinant.

   1. Identity Matrix - If matrix $A$ is the identity matrix $I$, $D(A) = \frac{|IB|}{|B|} = \frac{|B|}{|B|} = 1$.

   2. Row exchange - If two rows of $A$ are exchanged, those exact same corresponding rows are exchanged in $AB$. Since exchanging rows flips the sign of a determinant, so $|AB|$ changes sign. Hence $D(A)$ changes sign too.

   3. Linearity - If one row of $A$ multiplied by a scalar $t$, that same row in $AB$ is also multiplied by $t$. So $|AB|$ and $D(A)$ is multiplied by $t$. The same linear behavior holds true when adding rows together.

   $\therefore$ $\frac{|AB|}{|B|}$ has the same properties that define $|A|$, and $\frac{|AB|}{|B|} = |A|$.
   If $B$ is singular, then the product $AB$ is also singular. So $|AB| = 0$, and also $|A||B| = 0$, hence $|AB| = |A||B|$ still holds.

10. $|A^T| = |A|$. For a $2 \times 2$ matrix, $$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = \begin{vmatrix} a & c \\ b & d \end{vmatrix} = ad - bc$$ Because of this property, all the above properties apply to columns too. For example,

    • Swapping two columns changes sign of determinant.

    • Adding multiple of one column to another does not affect determinant.

    • Column of zeroes means determinant is 0.

6.2 Determinants as Area/Volume

6.2.0.1 Theorem

For a $2 \times 2$ matrix $A$, abs($|A|$) is the area of the parallelogram formed by the columns of $A$.
For a $3 \times 3$ matrix $A$, abs($|A|$) is the volume of the parallelepiped (a 3D slanted box) formed by the columns of $A$.
Proof.
Consider the $2 \times 2$ diagonal matrix, $$A = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$$ $\begin{bmatrix} a \\ 0 \end{bmatrix}$ represents a horizontal vector along the x-axis, $\begin{bmatrix} 0 \\ d \end{bmatrix}$ along the y-axis. These vectors are perpendicular, so the shape is a perfect rectangle. $$abs \left( \begin{vmatrix} a & 0 \\ 0 & d \end{vmatrix} \right) = abs(ad) = \text{Area of rectangle}$$ To transform any $2 \times 2$ matrix $A = [a_1\quad a_2]$ into a diagonal matrix without changing the absolute value of the determinant,

1. Swapping 2 columns - Flips the sign of $|A|$ but absolute value is unchanged.

2. 2D Shear, Replacing column $a_2$ with $a_2 + ca_1$ - Imagine some line $L$ passing through the origin and vector $a_1$. Vector $a_2$ dictates the top of the parallelogram. The line $a_2 + L$ passes through $a_2$ and is perfectly parallel to $L$. If we add $ca_1$ to $a_2$, the new vector $a_2 + ca_1$ simply slides along that top parallel line $a_2 + L$. Because the base $a_1$ hasn’t changed, and the points $a_2$ and $a_2 + ca_1$ have the exact same perpendicular distance to $L$ (the height is identical), both parallelograms calculate to the exact same base $\times$ height. Therefore, their areas are perfectly equal and $|A|$ is unchanged.

Similarly, for the $3\times 3$ diagonal matrix, $$A = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$$ The volume of the standard rectangular box is just $abs(|A|) = a \times b \times c$.
The same logic applies to transform any $3 \times 3$ matrix $A = [a_1 \quad a_2 \quad a_3]$.

• 3D Shear - The volume of a 3D parallelepiped is calculated as the area of its base multiplied by its height. Let the base be the flat 2D plane formed by vectors $a_1$ and $a_3$, known mathematically as $\text{Span}\{a_1, a_3\}$. The height is determined by how far vector $a_2$ sticks out from that base plane. The tip of $a_2$ lives in a parallel plane defined as $a_2 + \text{Span}\{a_1, a_3\}$. If we alter the matrix by replacing $a_2$ with $a_2 + ca_1$, that new vector just slides around inside that top parallel plane. Since the vector only moves sideways within the parallel plane, any vector $a_2 + ca_1$ maintains the exact same perpendicular height as the original $a_2$. With the base area identical and the height unchanged, the total volume of the parallelepiped is perfectly preserved when $\begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix}$ is changed to $\begin{bmatrix} a_1 & a_2 + ca_1 & a_3 \end{bmatrix}$.

6.3 Linear Transformations

6.3.0.1 Theorem

Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be a linear transformation determined by a $2 \times 2$ matrix $A$. Then for any parallelogram $S$, the area of the transformed shape $T(S) = abs(|A|) \times$ area of $S$.
3D space: For $T$ determined by a $3 \times 3$ matrix $A$, and $S$ a parallelepiped in $\mathbb{R}^3$, then volume of $T(S) = abs(|A|) \times$ volume of $S$.
Proof. First, imagine a parallelogram $S$ anchored at the origin, formed by two vectors $b_1$ and $b_2$. We can put these vectors into their own matrix, $B = \begin{bmatrix} b_1 & b_2 \end{bmatrix}$. Applying the transformation matrix $A$ to this shape, the new transformed shape $T(S)$ is defined by the matrix product $AB$. The new area is $abs(|AB|) = abs(|A||B|) = abs(|A|) \times abs(|B|) = abs(|A|) \times \text{area of } S$.
Then any arbitrary parallelogram can be written in the form $\mathbf{p} + S$, where $\mathbf{p}$ is a translation vector. $$\begin{align*} \text{area of }T(\mathbf{p} + S) &= \text{area of }(T(\mathbf{p}) + T(S)) \\ &= \text{area of} T(S) \\ &= abs(|A|) \times \text{area of }S \\ &= abs(|A|) \times \text{area of }\mathbf{p} + S \end{align*}$$ The above proof is similar for $3 \times 3$ case.

Expanding to Arbitrary Shapes
For example, stretching a unit circle to an ellipse.
$\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}$, representing the coordinates of the points forming the original unit circle, satisfying the circle equation $u_1^2 + u_2^2 = 1$.
Diagonal matrix $A$ represents the transformation that stretches the $u_1$ axis by $a$ and the $u_2$ axis by $b$, $$A = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$$ Multiplying $\mathbf{x} = A\mathbf{u}$, then $x_1 = au_1$ and $x_2 = bu_2 \implies$ $u_1 = x_1/a$ and $u_2 = x_2/b$. Thus we get the ellipse equation $\frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} = 1$.
As before, with $D$ representing the unit circle and $T$ the transformation, $$\begin{align*} \text{area of ellipse} &= \text{area of }T(D) \\ &= abs(|A|) \times \text{area of }D \\ &= ab \times \pi 1^2 \\ &= \pi ab \end{align*}$$